Category:Anderson Localisation

We will consider a line of $$N$$ coupled oscillators of mass 1. Let $$x_i$$ denote the displacement of the $$i$$th oscillator from the position it occupies when the system is minimum energy. Then the equations of motion of the system are given by,

$$ \dot{x_0} = 0 \,$$ $$ \ddot{x_i} = k_{i - 1} x_{i - 1} - (k_{i - 1} + k_i) x_{i} + k_{i} x_{i + 1} \,$$ $$ \dot{x_N} = 0 \,$$

where $$ k_i $$ is the elasticity of the spring connectiing oscillator $$ i $$ with oscillator $$ i + 1 $$. If we write

$$ z = \left( x_0, \dot{x_0}, x_1, \dot{x}_1, \cdots x_{N - 1}, \dot{x}_{N-1} \right)^{T} $$

The system of linear equations can be written as

$$ \dot{z} = \mathbf{A}z $$

where

$$ \mathbf{A} = \left(\begin{array}{ccccccccccc} 0 & 1 & 0 & 0 & 0  & 0 & \cdots & 0 & 0 & 0\\         -k_{N-1} - k_0 & 0 & k_0 & 0 & 0  & 0 & \cdots & 0 & k_{N-1} & 0\\        0 & 0 & 0 & 1 & 0  & 0 & \cdots & 0 & 0 & 0\\        k_0 & 0 & -k_0 - k_1 & 0 & k_1 & 0  & \cdots & 0 & 0 & 0\\        0 & 0 & 0 & 0 & 0  & 1 & \cdots & 0 & 0 & 0\\        0 & 0 & k_1 & 0 & - k_1 - k_2 & 0  & \cdots & 0 & 0 & 0\\        \vdots \\        0 & 0 & 0 & 0 & 0  & 0 & \cdots & 0 & 0 & 1\\        k_0 & 0 & 0 & 0 & 0  & 0 & \cdots & 0 & -k_{N-2} - k_{N-1} & 0 \end{array} \right).$$

The matrix $$ \mathbf{A} $$ has $$ N -1 $$ linearly independent eigenvectors (proof: With $$N $$ oscillators and the constraint $$ \sum_i \dot{x}_i = 0 $$, there are $$ N-1 $$ degrees of freedom.) They correspond to imaginary eigenvalues (Proof: It must be the case that they are imaginary as the system being described conserves energy, and so no non-imaginary eigenvalues are possible. Can I show it mathematically?) The solution can thus be written as a superposition of periodic eigensolutions.

Localization
When the elasticity of all of the springs is the same, the eigensolutions corresponding to eigenvalues $$\omega_Q$$ are given by,

$$ u_n^Q(t) = \frac{1}{\sqrt{N}} e^{i(Qna - \omega_Q t)}, $$

where $$ Q = 2 \pi \frac{n}{N} $$, $$ n \in \mathbb{Z} \cap [-\frac{N}{2},\frac{N}{2}] $$ and $$ \omega_Q = 2\sqrt{k}\sin(Q/2) $$. In this case, the magnitude of the eigenvector at oscillator $$n $$, $$ |u_n^Q| $$, is $$\frac{1}{\sqrt{N}} $$, specifically it is independent of $$n$$. We thus see that the energy is equally distributed amongst all the nodes. We define the ``participation ratio" of the eigenstate corresponding to $$Q$$, $$p_Q$$ as

$$ p_Q = 1 / \sum_{n = 0}^{N-1}{|u_n^Q|^4}. $$

We see that for constant elasticity, $$ p_Q $$ is $$ \frac{1}{N} $$. If in the otherhand, the eigenvector were a delta function, so that all the energy is located on one node, the participation ratio would be $$ 1 $$. It is interesting to consider how the participation ratio changes as the elasticity of the springs becomes less uniform. The phenomenon whereby the energy becomes more localised as the elasticity becomes more variable is known as Anderson localisation.

Simulations
There are two ways to proceed with the computations. The first way is to implement a finite difference approach:

$$ u_n(t + dt) = 2 u_n(t) - u_n(t - dt) + F_n(t) dt^2 $$

where $$ F_n(t) = k_{n-1} (x_n(t) - x_{n-1}(t)) + k_n ( x_{n+1}(t) - x_n(t) ) $$.

The second approach is to find the eigenvalues and eigenvectors of the matrix $$ \mathbf{A} $$ above directly, and hence find the explicit solution to the problem for all time.

I have implemented the finite difference scheme described above. The resulting figures can be viewed at http://dl.dropbox.com/u/6360844/coupled_oscillators_JN.pdf in pdf format, we cannot figure out how to upload them directly.